# If a spring is cut in half, does the period change?

1. Nov 23, 2008

### clairez93

1. The problem statement, all variables and given/known data

A simple harmonic oscillator consists of a mass m and an ideal spring with spring constant k. Particle oscillates as shown in (i) with preriod T. If the spring is cut in half and used with the same particle, as shown in (ii), the period will be.

A) 2T
B) $$\sqrt{2}$$T
C) T/$$\sqrt{2}$$
D) T
E) T/2

2. Relevant equations

T = 2$$\pi$$$$\sqrt{m/k}$$

3. The attempt at a solution

I figured that since the length of the spring isn't relevant to the period, the period would stay the same. The answer provided is C. I'm not sure why.

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2. Nov 23, 2008

### clairez93

Hm, a thought, does cutting the spring in half alter the constant in some way?

3. Nov 23, 2008

### clairez93

This is all I can think of, because the mass stays the same, therefore the only thing that would change is the K, however why would cutting the spring in half change the K?

4. Nov 23, 2008

### clairez93

I read this problem in an archive, however, the answers given there don't seem to make sense. They start talking about the amplitude changing, however, isn't the amplitude independent of the period?

5. Nov 23, 2008

### lonewolf5999

Consider a spring which extends by length x under a load of mass m.
Connect two such springs in series to form one long spring. What would be the new extension of this spring under a load of mass m? What is the effective new spring constant of this long spring?
Can you extend this idea to cutting a spring in half?

6. Nov 24, 2008

### clairez93

The extension would be longer, but how does that affect the spring constant?

7. Nov 24, 2008

### Staff: Mentor

If you take the full spring and hang a weight W from it, it will stretch an amount Δx. So the spring constant is k = W/Δx.

Now consider the two halves of that spring. How much stretch does each half get? The force (W) is the same, so what must be the spring constant for each half?

8. Nov 24, 2008

### clairez93

Oh I see. So the first spring:
$$k = mg/x$$

Thus, $$T = 2\pi\sqrt{x/g}$$

So in the second spring:

$$k = 2mg/x$$

Thus, $$T' = 2\pi\sqrt{x/2g} = T/\sqrt{2}$$

Is this correct?

9. Nov 24, 2008

### Staff: Mentor

Yes, that's right. The key thing is that the spring constant of half a spring is twice the spring constant of the full spring. (This should make a bit of sense. You'd expect it to be harder to pull a shorter spring to the same extension.) And since the period depends on the spring constant, it will change accordingly.